Here are my thoughts on the problems we solved from this year's electronic round of Challenge 24.

## A. Safe

I didn't get a complete solution to this, but I was able to solve cases 1-9. I noticed that in most cases the state space for the wheels (keys, but the pictures made me think of them as wheels) is reasonably small. For each password, the solution will involve first pressing the keys a number of times to set up the initial state, and then typing the password. So we want to find the initial states from which we can input the password, and pick the cheapest amongst them.

This can be done by dynamic programming, running backwards over the password. When there are no more characters to type, any state is fine. Considering each letter in turn, we need to find the states in which we can type the next letter on one of the keys, and from which the resulting state allows us to complete the password.

This can be done by dynamic programming, running backwards over the password. When there are no more characters to type, any state is fine. Considering each letter in turn, we need to find the states in which we can type the next letter on one of the keys, and from which the resulting state allows us to complete the password.

For A10 there are 6.4 billion states, and it was going to take a long time (it did churn out the first two passwords in about 3 hours). I suspect it might be possible to do it by keeping an explicit list/set of reachable states and working from there, rather than iterating over the entire state space.

## B. Wiretapping

This problem basically just requires an application of Kirchhoff's Theorem, which gives the number of spanning trees of the graph in polynomial time. To count the number of spanning trees that use the tapped link, just collapse the two endpoints into a single vertex and count the number of spanning trees of the remaining graph. The ratio between the number of spanning trees in this reduced graph and the original graph is the answer.

One complication is that the number of spanning trees may overflow a double-precision float. I worked around that by scaling down the Laplacian matrix by some factor f (I used 1/32 for the larger cases), which reduces the determinant of the cofactor by f

One complication is that the number of spanning trees may overflow a double-precision float. I worked around that by scaling down the Laplacian matrix by some factor f (I used 1/32 for the larger cases), which reduces the determinant of the cofactor by f

^{N-1}.## C. Visual Programming - VM

This was a fairly straightforward implementation problem, just requiring careful interpretation of the rules, and recognising that a lot of the information given is unnecessary (e.g., for lines you only need to keep the two endpoints).

## D. Visual Programming

Due to lack of time I only solved D1, D2 and D4, and had a broken solution to D3. As with most VM problems, I invented a slightly higher-level language that is possible (but still hard) to code in by hand, and write a simpler compiler. I made a few simplifying assumptions

- Polygons are laid out one below the other. Each polygon is rectangular, except that the right hand edge may be ragged.
- Each line is only ever used in one direction. To ensure this, the outgoing lines from a polygon all emanate from the top part of the right edge, and incoming lines arrive at the bottom part of the right edge. There is also an unconditional jump from each polygon to the next one to ensure that the routing terminates before encountering any of the incoming lines.
- To avoid lines cutting through polygons, each line emanates directly horizontally from the polygon to (256, y), then has a second segment to the target point. The x value of the start is chosen depending on which register is to be used in the comparison.
- I reserved registers 0 and 1 to always contain the values 0 and 1, which made it easier to construct some other operations, as well as to build an unconditional jump.

The programming mini-language just specifies instructions and conditional a list of conditional jumps that takes place after each instruction (falling through to the next instruction if none match). The instructions I wrote at the level of setting the parameters (A-F, X-Z, T, Q, L, H), but I had useful defaults for all of them so that most instructions only changed one or two defaults e.g. setting A=3 B=15 would generate an instruction that copies from SIO into register 3.

Once all that is done, it is not particularly difficult for anyone who has done any assembly to write code for D1, D2, and D4. The only slightly tricky part was to extract the lower 8 bits from the result, which I did by multiplying to 256 then dividing by 256.

## G. Spy Union

At first I thought this could be done greedily, but it doesn't work: reinstating a fired person might let you fire an additional person

*in each hierarchy*. Instead, we can use network flow. Let's consider just one hierarchy for now, and connect the source to every node, connect each node to its parent, and connect the root to the sink (directed edges). The edges from the source have capacity 1, with the interpretation that they have flow if the corresponding employee is fired. The flow through each intermediate edge will then be the total number of employees fired from the corresponding subtree, so we will set the capacities to be the maximum number we can fire each subtree. The maximum flow will thus be the maximum number of people we can fire.
However, we have two hierarchies. Since network flow is symmetric, we can apply the same treatment to the second hierarchy, but connect the source to the root, and the individual employees to the sink. To join up the problems, we get rid of the source in the first graph and the sink in the second graph, and create capacity-1 edges directly from each employee in the second graph to the matching employee in the first. Once again, these edges have flow if the corresponding employee is fired, and the maximum flow gives the solution.